Let OABCD be a pentagon in which the sides OA and CB are parallel and the sides OD and AB are parallel as shown in figure. Also, OA:CB=2:1 and OD:AB=1:3. if the diagonals OC and AD meet at x, find OX:XC.

Answer

`2:5`

Explanation

Let O be the origin of reference.
Let the position vectors of A,B,C and D be a,b,c and d,
respectively.
then, `OA:CB=2:1`
`implies (OA)/(CB)=(2)/(1)implies OA+2CB`

`implies OA=2CB`
`implies a=2(b-c)` . . . (i)
and `OD:AB=1:3`
`implies (OD)/(AB)=(1)/(3) implies 3OD=AB`
`implies 3OD=AB`
`implies 3d=(b-a)=b-2(b-c)` [using eq. (i)]
`implies 3d=-b+2c` . . . (ii)
Let OX:XC=`lamda:1 and AX:XD=mu:1`
Now, X divides OC in the ratio `lamda:1`. therefore,
PV of `X=(lamdac)/(lamda+1)` . . . (iii)
X also divides AD in the ratio `mu:1`
PV of X`=(mud+a)/(mu+1)`
from eqs. (iii) and (iv), we get
`(lamdac)/(lamda+1)=(mud+a)/(mu+1)`
`implies ((lamda)/(lamda+1))c=((mu)/(mu+1))d+((1)/(mu+1))a`
`implies((lamda)/(lamda+1))c=((mu)/(mu+1))((-b+2c)/(3))+((1)/(mu+1))2(b-c)`
[using eqs. (i) and (iv)]
`implies((lamda)/(lamda+1))c=((2)/(mu+1)-(mu)/(3(mu+1)))b+((2mu)/(3(mu+1))-(2)/(mu+1))c`
`implies ((lamda)/(lamda+1))c=((6-mu)/(3(mu+1)))b+((2mu-6)/(3(mu+1)))c`
`implies ((6-mu)/(3(mu+1)))b+((2mu-6)/(3(mu+1))-(lamda)/(lamda+1))c=0`
`implies (6-mu)/(3(mu+1))=0 and (2mu-6)/(3(mu+1))-(lamda)/(lamda+1)=0`
(since, b and c are non-collinear)
`implies mu=6 and lamda=(2)/(5)`
Hence, `OX:XC=2:5`